The solution of the differential equation $\left(x y^4+y\right) d x-x d y=0$, is

$3 x^4 y^3+4 x^3=C y^3$

We have,

$\left(x y^4+y\right) d x-x d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{x y^4+y}{x}$

$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=y^4$

$\Rightarrow \frac{1}{y^4} \frac{d y}{d x}+\left(\frac{-1}{y^3}\right) \frac{1}{x}=1$

Let $-y^{-3}=v$. Then, $3 y^{-4} \frac{d y}{d x}=\frac{d v}{d x}$

∴  $\frac{1}{3} \frac{d v}{d x}+\frac{v}{x}=1 \Rightarrow \frac{d v}{d x}+\frac{3}{x} v=3$           ....(i)

This is a linear differential equation with integrating factor $x^3$

Multiplying both sides of (i) by $x^3$ and integrating, we get

$v x^3 =\frac{3 x^4}{4}+C$

$\Rightarrow \frac{-x^3}{y^3}=\frac{3 x^4}{4}+C$

$\Rightarrow -4 x^3=3 x^4 y^3+4 y^3 C$

$\Rightarrow 3 x^4 y^3+4 x^3=-4 C y^3 \Rightarrow 3 x^4 y^3+4 x^3=\lambda y^3$